can be checked that all states have period 1, simply because p
i,i
> 0 f or all i = 1, 2, 3.
(d) Irreducible? YES because there is a path from every state to any other state.
Aperiodic? NO because the times n for which p
(n)
1,1
> 0 are 2, 4, 6, . . . and their gcd is
2.
(e) Ir reducible? YES because there is a path from every state to any other state.
Aperiodic? YES because the times n for which p
(n)
1,1
> 0 are 1, 2, 3, 4, 5, . . . and their
gcd is 1.
8.
Consider the knight’s tour on a chess board: A knight selects one of the next positions
at random independ ently of the past.
(i) Why is this process a Markov chain?
(ii) What is the state space?
(iii) Is it irreducible? Is it aperiodic?
(iv) Find the stationary distribution. Give an interpretation of it: what does it mean,
physically?
(v) Which are the most likely states in steady-state? Which are the least likely ones?
Solution. (i) Part of the problem is to set it up correctly in mathematical terms.
When we say that the “knight selects one of the next positions at random indepen-
dently of the past” we mean that the next position X
n+1
is a function of the current
position X
n
and a random choice ξ
n
of a neighbou r. Hence the problem is in the same
form as the one above. Hence (X
n
) is a Markov chain.
(ii) The state space is the set of the squares of the chess board. There are 8 × 8 = 64
squares. We can label them by a pair of integers. Hence the state space is
S = {(i
1
, i
2
) : 1 ≤ i
1
≤ 8, 1 ≤ i
2
≤ 8} = {1, 2, 3, 4, 5, 6, 7, 8} ×{1, 2, 3, 4, 5, 6, 7, 8}.
(iii) The best way to see if it is irreducible is to take a knight and move it on a chess
board. You will, indeed, realise that you can fi nd a path that takes the knight from
any square to any other square. Hence every state communicates with every other
state, i.e. it is irreducible.
To see what the period is, fin d the perio d for a s pecific state, e.g. from (1, 1). You
can see that, if you start the knight from (1, 1) you can return it to (1, 1) only in
even number of steps. Hence the period is 2. So the answer is that the chain is not
aperiodic.
(iv) You have no chance in solving a set of 64 equations with 64 unknowns, unless you
make an educated guess. First, there is a lot of sym metry. So squares (states) that
are symmetric with respect to the centre of the chess board must have the pr ob ability
under the stationary d istrib ution. So, for example, states (1, 1), (8, 1), (1, 8), (8, 8)
have the same probability. And so on. Second, you sh ould realise that (1, 1) must be
less likely than a square closer to the centre, e.g. (4, 4). The reason is that (1, 1) has
fewer next states (exactly 2) th an (4, 4) (which has 8 next states). So let us make the
guess that if x = (i
1
, i
2
), then π(x) is proportional to the number N (x) of the possible
next states of the squ are x:
π(x) = CN(x).
But we must SHOW that this choice is correct. Let us say that y us a NEIGHBOUR
of x if y is a possible next s tate of x (if it is possible to move the kn ight from x to y
7